Energy & Momentum: Ski Jumping Homework

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The discussion revolves around solving a ski jumping physics homework problem involving energy and momentum conservation. The ski jumper, with a mass of 70 kg, descends a hill with an initial velocity of 9 m/s and a radius of curvature of 8 m. Key calculations include determining the jumper's velocity at takeoff (5.84 m/s), the maximum height achieved (4.2 m), and the impulse experienced while on the hill (-290 Ns). The conversation emphasizes the importance of using both energy and momentum equations to analyze the system accurately, particularly when considering the interaction between the jumper and a moving hill of mass 1000 kg.

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Hi, I got homework and I need someone to check it, please. :)
http://shrani.si/f/1G/JS/1UO1SWSw/skakalnica.jpg

Ski jumper (m=70kg) goes down the hill with v0=9m/s. The hill has got R (radius of curvature = 8m). There is no friction (the hill is 1m above Earth - like on picture).
1. a) What is the velocity when the ski jumper just jumps?
My answer: Wk(begining)+Wp(begining)=Wk(end)+Wp(end)=
=(70*9^2)/2+70*10*1=(7*vfinal^2)/2+70*10*3,34 (h0+h)=
vfinal=5,84m/s

b) Find the highest point!
hmax=v0y^2/(2g)=(5,84*cos45)^2/20=0,85+h0=4,2m

c) What is the impulse of the hill to the jumper (vertical) when he is on the hill?
∫Fydt=G(end)-G(begining)
∫Fydt=0-m*v(end)y=0-70*sin45=-290Ns

2. Let the hill move without friction. (Mhill=1000kg) When jumper hits the hill (v0=9m/s) it is at rest.

a) Does the energy of the system (jumper+hill) preserve? Prove with Work of outer and inner forces.
I assume it does, but don't know how to prove it.?!

b) What is the velocity (to the observer on earth) when jumper leaves the hill? At what angle does it point?
I am not sure here, whether I use the conservation of energy equation or mumentum equation?:
m*g*h0+(m*v0^2)/2=mgh1+(m*v1^2)/2+(Mhill*v1^2)/2=
=1,5m/s angle=45°


or:
m*v0=(m+Mhill)v1=0,589m/s.

c) What is the velocity of the hill?
Is it here possible that I use here the equation of mumentum to get this vx and that in (b) I use energy conservation to get v=1,5. And get angle (in b) like cosθ=v/v1...θ=67°

Thanks for help. :)
 
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gasar8 said:
vfinal=5,84m/s
Seems a bit low. If you keep getting that answer, please post all your working.
b) Find the highest point!
hmax=v0y^2/(2g)=(5,84*cos45)^2/20=0,85+h0=4,2m
The trouble with using numerical intermediate results is accumulation of errors. It's best to keep everything purely symbolic (algebraic) until the final step of each answer. It also makes it easier to spot errors.
b) What is the velocity (to the observer on earth) when jumper leaves the hill? At what angle does it point?
I am not sure here, whether I use the conservation of energy equation or mumentum equation?:
You now have two unknowns: the velocity of the jumper (at leaving) and the velocity of the hill. So you will need two equations. Work energy is certainly one. You can use momentum if there are no forces external to the system in the direction of the momentum you calculate.
 
Last edited by a moderator:
Hi, thank you for your quick reply!

a) From energy equation that I used I get v=5,92m/s but not more. Also the same from v^2=v0^2-2gh...

b) When I put this new result in I get 4,23m for max height.

2. b) I have no clue how to solve this. So, the energy equation is right? Is this velocity that I get the velocity of the jumper and the hill to the right side (x axis)?
Can you give me some hints please? :)
Thank you.
 
gasar8 said:
a) From energy equation that I used I get v=5,92m/s but not more. Also the same from v^2=v0^2-2gh...

b) When I put this new result in I get 4,23m for max height.
Hmm.. I agree with your answer now. Don't know what I did wrong before. Sorry about that.
2. b) I have no clue how to solve this. So, the energy equation is right? Is this velocity that I get the velocity of the jumper and the hill to the right side (x axis)?
Can you give me some hints please? :)
Thank you.
Your energy equation assumes the hill and the jumper will have the same speed at take-off. That cannot be true.
To keep it simple, let's allow a third unknown. You have the horizontal speed of the hill, the vertical speed of jumper at take-off and horizontal speed of jumper at take-off. That last can be either relative to the hill or relative to the ground - your choice - as long as you are consistent.
In addition to the energy equation and horizontal momentum equation, you should be able to find an equation relating those three speeds to the slope of the hill at the take-off point.
You understand why momentum of the skier+hill system is conserved horizontally but in no other direction, right?
 
Ufff...just went trough maths and I really hope it is right. :D I have to hand this homework in tomorrow.

So for 2.b)
Momentum is conserved in x axis. So m1*v0+0=m1*vx1+m2*v2. From this equation I get v2=(m1*(v0-vx1))/m2.
This vx1 is the x component of skiers velocity!
Then I used energy equation (kinetic and potential energy of skier+hill in the beginning is equal to the kinetic and potential energy in the end). So:
m1*v0^2/2 + 0 + m1*g*h0 + 0 = m1*v1^2/2 + m2*v2^2/2 + m1*g*h1 + 0 (This v1 is the whole velocity not only in x component).
Then I put momentum v2 into this equation and get the equation to the power 4, hehe. Difficult to not do any mistake. :D And got v1^2=27,28 ==>v1=5,22m/s. When I put this into momentum v2 (again *cos45) I get v2=0,37m/s.
Those results seems quite intuitive to me (v1 is a little bit less then when the hill is at rest and v2 is also small because of its mass).
Please tell me this is right, hehe. ;D
Now I only wonder how to prove (with work of forces) that energy is conserved?

Thank you a lot! Have a nice day!
 
gasar8 said:
So for 2.b)
Momentum is conserved in x axis. So m1*v0+0=m1*vx1+m2*v2. From this equation I get v2=(m1*(v0-vx1))/m2.
This vx1 is the x component of skiers velocity!
Then I used energy equation (kinetic and potential energy of skier+hill in the beginning is equal to the kinetic and potential energy in the end). So:
m1*v0^2/2 + 0 + m1*g*h0 + 0 = m1*v1^2/2 + m2*v2^2/2 + m1*g*h1 + 0 (This v1 is the whole velocity not only in x component).
That all looks ok.
Then I put momentum v2 into this equation and get the equation to the power 4, hehe. Difficult to not do any mistake. :D And got v1^2=27,28 ==>v1=5,22m/s. When I put this into momentum v2 (again *cos45) I get v2=0,37m/s.
Not sure how you got to there (not saying it's wrong). You had two equations and three unknowns: v1x, v1y, v2. What was your third equation?
Those results seems quite intuitive to me (v1 is a little bit less then when the hill is at rest and v2 is also small because of its mass).
Please tell me this is right, hehe. ;D
Now I only wonder how to prove (with work of forces) that energy is conserved?

Thank you a lot! Have a nice day!
Yes, the answer certainly looks reasonable.
 
I was thinking that if I get v1x I don't need v1y, because I know the angle? So v1=v1x/cos45?
I spotted now one mistake (in previous reply I calculated v1=v1x*cos45, but it is devided by cos45). So now I get v1=4,18m/s and v2=0,42m/s.
 
gasar8 said:
I was thinking that if I get v1x I don't need v1y, because I know the angle? So v1=v1x/cos45?
No. Your v1x is relative to the ground. The relationship between horizontal and vertical motion of the skier that's determined by the hill slope will be for speeds relative to the hill.
That said, it should only introduce a small error.
 

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